If the radioactive isotope strontium $240$ has a half life of $120$ years, how long until it decays to only $60\%$ of its original radioactivity?

Question

I been trying to solve this problem for hours and the only thing i came up with, was a formula for their relationship.

$1/2A = A_0 e^{120r}$

$\ln(1/2 = e^{120}r)$

$\ln(1/2) = 120r$

$r = \ln(0.5)/120 $

$r = -0.0057762265$

So, when the isotope is $120$ years old the percentage is $-0.0057762265$. But I don't know how to find how long until the isotope decays to $60\%$? I really need on understanding this problem, I don't only want the answer I want an explanation of how to do it? please help, I have an exam tomorrow base on problems like this.

Answer 1

You know that the basic exponential growth/decay equation is

\begin{equation} A=A_0e^{rt} \end{equation}

You are told that when $t=120$ that

\begin{equation} A=\tfrac{1}{2}A_0e^{120r} \end{equation}

which you solved correctly for $r$.

Now you wish to know the value of $t$ which causes

\begin{equation} 0.60A_0=A_0e^{-0.005762265\,t} \end{equation}

so you must solve for $t$ the following equation:

\begin{equation} e^{-0.005762265\,t}=0.60 \end{equation}

which you should have no trouble doing.

Answer 2

Your approach is the hard way. Do it like this:

Every time $120$ years pass, you multiply the amount by $1/2$.

If $t$ is the number of years that have passed, then $t/120$ is the number of $120$-year periods that have passed. That is therefore how many times you multiply by $1/2$. Hence the amount remaining after $t$ years is $$ \text{original amount} \times \left( \frac 1 2 \right)^{t/120}. $$ So you need to know what $t$ is when $\left(\dfrac 1 2 \right)^{t/120}\!\!\!\!\! = 0.6$.

You have $$\frac t {120} = \log_{1/2} 0.6.$$ And then you can find $t$.

The reason for bringing in base-$e$ logarithms and base-$e$ exponential functions is to take about instantaneous rates of change, i.e. derivatives. As long as the problem doesn't involve those, then bringing in the number $e$ is just a pointless complication.

However, by your method one should say $$ A = A_0 e^{rt} $$ $$ \frac 1 2 = e^{r120} $$ $$ r = \text{a particular negative number} $$ $$ e^{rt} = 0.6 $$ $$ rt = \ln 0.6 \quad \text{(This is negative.)} $$ $$ t = \frac{\ln 0.6} r \quad \text{(This is positive.)} $$

One thing that happens when this method is used is that rounded values are used where exact values are available. The happens especially when a student see something like $e^{t\ln 0.7}$ and instead of realizing that that is exactly $0.7^t$, gets a rounded value of $0.7$ from a calculator and goes on from there.